If the $n^{t h}$ term of an A.P. be $(2 n - 1),$ then the sum of its first $n$ terms will be

n^{2} - 1

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(n - 1)^{2} + (2 n - 1)

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(n - 1)^{2} - (2 n - 1)

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n^{2}

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Solution

The correct options are
B $(n - 1)^{2} + (2 n - 1)$
C $n^{2}$
Given,
$\Rightarrow n^{t h}$ term of AP $= 2 n - 1$

$\Rightarrow 1^{s t}$ term of AP $= 2 (1) - 1 = 1$
$\Rightarrow 2^{n d}$ term of AP $= 2 (2) - 1 = 4 - 1 = 3$
$\Rightarrow$ Common difference $(d) =$ $2^{n d}$ term - $1^{s t}$ term $= 3 - 1 = 2$

$\Rightarrow$ Sum of first $n$ terms of AP $= \frac{n}{2} [2 a + (n - 1) d]$

$\Rightarrow$ Sum $=$ $\frac{n}{2} [2 \times 1 + (n - 1) 2]$
$=$ $\frac{n}{2} \times (2 n)$

$=$ $n^{2}$

$=$ $n^{2} + (2 n + 1) - (2 n + 1)$ (adding and subtracting $2 n + 1$ )

$=$ $(n^{2} - 2 n + 1) + 2 n - 1$

$=$ $(n - 1)^{2} + (2 n - 1)$

Correct options $: B, D$

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